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3 years ago

11

How is capacitance related to the distance between the plates of a capacitor?

1 answer:

svetlana [45]3 years ago

5 0

Answer:C

Explanation: I studied, and C is correct

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A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is

Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

$\frac{\Delta l}{l_{o}}=\frac{F}{AY}$ (1)

$T=2 \pi \sqrt{\frac{l_{o}}{g}}$ (2)

Where:

$\Delta l=0.05 mm=5(10)^{-5} m$ is the length the steel wire streches (taking into account 1mm=0.001 m)

$l_{o}$ is the length of the steel wire before being streched

$F=mg=(2 kg)(9.8 m/s^{2})=19.6 N$ is the force due gravity (the weight) acting on the pendulum with mass $m=2 kg$

$A$ is the transversal area of the wire

$Y=2(10)^{11} Pa$ is the Young modulus for steel

$T$ is the period of the pendulum

$g=9.8 m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin by finding $A$ :

$A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4}$ (3)

Where $d=1.1 mm=0.0011 m$ is the diameter of the wire

$A=\pi \frac{(0.0011 m)^{2}}{4}$ (4)

$A=9.5(10)^{-7}m^{2}$ (5)

Knowing this area we can isolate $l_{o}$ from (1):

$l_{o}=\frac{\Delta l AY}{F}$ (6)

And substitute $l_{o}$ in (2):

$T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}}$ (7)

$T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}}$ (8)

Finally:

$T=1.39 s$

3 0

3 years ago

a book is sitting still on your desk. are the forces acting on the book balanced or unbalanced? explain.

GrogVix [38]

<span>They are balanced. If the forces were not balanced the book would move*. In this example, the downward force of gravity on the book is counterbalanced by the upthrust of the desk. </span>

8 0

3 years ago

Read 2 more answers

You are a particle physicist at the Large Hadron Collider who is tasked with designing an apparatus to separate annihilation pro

JulijaS [17]

Find solution in the attachments

4 0

3 years ago

Can someone help me?

Alik [6]

Answer:

Resultant = 3.05N

Explanation:

$r = \sqrt{{5}^{2} + 5^{2} - 2(5)(5) \cos(120) }$

$r = \sqrt{25 + 25 - 50(0.8142)}$

$r = \sqrt{50 - 40.71} = \sqrt{9.29}$

$r = 3.05$

6 0

3 years ago

A 2,000 kg car starts from rest and coasts down from the top of a 5.00 m long driveway that is sloped at an angel of 20o with th

ehidna [41]

Answer:

The speed at the bottom of the driveway is3.67m/s.

Explanation:

Height,h= 5sin20°= 1.71m

Potential energy PE=mgh= 2000×9.8×1.71

PE= 33516J

KE= PE- Fk ×d

0.5mv^2= 33516 - (4000×5)

0.5×2000v^2= 33516 - 20000

1000v^2= 13516

v^2= 13516/1000

v =sqrt 13.516

v =3.67m/s

5 0

3 years ago